Question

If (p^2+1)(q^2+1) + 36 = 12(p+q);
then (p^3+q^3) =
11:11 am​

Answer 1

Given : (p² +1)(q² + 1)  + 36  = 12(p + q)

To Find : p³  + q³

Solution:

(p² +1)(q² + 1)  + 36  = 12(p + q)

=> p²q² + p² + q² + 1  + 36 = 12p + 12q

=>  p² + q²  + 36   -12p - 12q    + p²q² + 1 = 0

Adding and subtracting   2pq

=>  p² + q²  + 36   -12p - 12q  + 2pq  + p²q² + 1  - 2pq= 0

(p + q - 6)²   = p² + q² + 36   -12p - 12q + 2pq

(pq - 1)² = p²q² + 1  - 2pq

=> (p + q - 6)²  +  (pq - 1)²  = 0

=> p + q - 6 = 0  and pq - 1 = 0

=> p + q   = 6     and pq  =  1

p + q   = 6  

cubing both sides

=> p³ + q³ + 3pq(p + q)   = 6³

=>  p³ + q³ + 3(1)(6)   = 216

=> p³ + q³  = 198

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Answer 2

SOLUTION

GIVEN

$\sf{( {p}^{2} + 1)( {q}^{2} + 1) + 36 = 12(p + q)}$

TO DETERMINE

The value of

$\sf{ {p}^{3} + {q}^{3} }$

EVALUATION

$\sf{( {p}^{2} + 1)( {q}^{2} + 1) + 36 = 12(p + q)}$

$\sf{ \implies \: {p}^{2} {q}^{2} + {p}^{2} + {q}^{2} + 1+ 36 = 12(p + q)}$

$\sf{ \implies \: {p}^{2} + {q}^{2} +12(p + q) - 2pq + 36 + {p}^{2} {q}^{2} - 2pq + 1= 0}$

$\sf{ \implies \: {(p + q - 6)}^{2} + {(pq - 1)}^{2} = 0}$

We know if the sum of the squares of two real numbers are zero then they are separately zero

$\sf{ {(p + q - 6)}^{2} = 0 \: \: and \: \: {(pq - 1)}^{2} = 0}$

$\sf{ \implies \: {(p + q - 6)} = 0 \: \: and \: \: {(pq - 1)} = 0}$

$\sf{ \implies \: p + q = 6 \: \: and \: \:pq = 1}$

$\sf{ {p}^{3} + {q}^{3} }$

$\sf{ = {(p + q)}^{3} - 3pq(p + q) }$

$\sf{ = {(6)}^{3} - 3 \times 6 \times 1 }$

$\sf{ = 216 - 18 }$

$\sf{ = 198 }$

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